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3 years ago

Why are three partial product in 1 (a) and only two partial product s in 1 (b)?

Mathematics

2 answers:

kakasveta [241]3 years ago

6 0

They are the same.

Please let me know if you need anymore help.

KIM [24]3 years ago

3 0

because partial 1(a) and partial 1(b) are the same.

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What is the sum of 90 minutes and 50 minutes? 130 minutes 1 hour, 40 minutes 2 hours 2 hours, 20 minutes

LiRa [457]

90 mins + 50 mins = 140 mins
60 mins ==1 hr 2*60 = 120 mins
140 mins -120 mins= 20 mins
2 hrs ans 20 mins is the answer

7 0

3 years ago

Read 2 more answers

I need your help on this one pls its another one but different question<br> and please dont guess

Ainat [17]

Answer:

All except the second one (angle g is not similar to angle r)

Step-by-step explanation:

Just look at the order of the letters. It it's the same, its similar.

4 0

3 years ago

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

igor_vitrenko [27]

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

$m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7$

So, the solution of the equation is

$y = Ae^{m_{1} t} + Be^{m_{2} t}$

where m₁ = 3 and m₂ = -7.

So,

$y = Ae^{3t} + Be^{-7t}$

Also,

$y' = 3Ae^{3t} - 7e^{-7t}$

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

6 0

4 years ago

Please I really need help 100 points

Afina-wow [57]

Multiply the first equation by 3 and the second equation by 2:

$\begin{cases}6x-9y=27\\6x+10y=8\end{cases}$

Subtract the two equation to get rid of the x variable:

$-9y-10y=27-8 \iff -19y = 19 \iff y = -1$

Use the value of y to determine the value of x: for example, using the first equation we have

$2x+3=9 \iff 2x=6 \iff x=3$

4 0

3 years ago

Read 2 more answers

I need help pls asap!!

Vera_Pavlovna [14]

Answer:

No, Electric Pole is 40 ft

Step-by-step explanation:

tan 66.8° = $\frac{x}{15}$

x = 35

but that is from the top of Danny, so you need to add 5 (his height)

Electric Pole is 40 ft

4 0

3 years ago

Why are three partial product in 1 (a) and only two partial product s in 1 (b)?​

Why are three partial product in 1 (a) and only two partial product s in 1 (b)?